Question

Given f(x)=ax/(x^2+b) find a and b such that f'(1)=0 and f'(0)=1.

Answer 1

First step is to take the first derivative of f(x). What do you get?

Answer 2

using quotient rule i got: x(x^2+b) - ax(2x) and all that divided by (x^2+b)^2

Answer 3

Hmm, that's not what I get...

Answer 4

oh i typed soemthing wrong, i actually got a(x^2+b) and then the rest

Answer 5

\[\frac{a(x^2+b) -ax(2x)}{(x^2+b)^2}\]?

Answer 6

yes

Answer 7

Okay. now can you evaluate that for x=0?

Answer 8

That will give you f'(0).

Answer 9

i get ab/b^2 so 0= a/b ?

Answer 10

no, f'(0) = a/b = 1

Answer 11

f'(0) = 1 from problem statement

Answer 12

now we need to evaluate f'(1) and set it equal to 0...

Answer 13

OH ok i see so a=b

Answer 14

right. that makes it easier :-)

Answer 15

th

Answer 16

as pretty simple lol, thanks :)

Answer 17

well, we aren't done yet! you don't have a value for a and b :-)

Answer 18

is a and b equal to one?

Answer 19

i tried f'(1) = 0 and subbed in 1 for x and made b=a

Answer 20

Yep, a = b = 1 is the solution. You have to be a little careful and check both equations, though. There are values of a and b that make f'(0) = 1 but fail on f'(1). 3, for example.

Answer 21

i see, thanks so much :)

Answer 22

I'm not saying that a= b=1 is incorrect, just that your check wasn't sufficient to be sure of a correct answer.

Answer 23

Want to do that other problem that you posted and closed?

Answer 24

lol i figured what i did wrong, but thanks anyways

Answer 25

no problem, no problem :-)